A deceased person is found by the coronoer to have a cocaine concentration of 50 ng/dL. What is the concentration of cocaine in his blood two hours earlier?

What is the rate? You must have that in your information somewhere.

rate is 1.1 ng/dL-hr

Is this a first order reaction? If so then

ln(No/N) = kt
Substitute No and solve for it.
N = 50
k = 1.1
t = 2.
If this is not a first order decay please repost giving ALL of the information in the problem.

To determine the concentration of cocaine in the deceased person's blood two hours earlier, we need to consider the concept of the body's elimination half-life for cocaine.

Cocaine has an elimination half-life of approximately 1 hour. This means that every hour, the concentration of cocaine in the body decreases by half.

Using this information, we can calculate the concentration of cocaine in the person's blood two hours earlier by working backwards from the given concentration of 50 ng/dL.

1. Start with the given concentration: 50 ng/dL.

2. Since the elimination half-life of cocaine is 1 hour, and we need to go back 2 hours, we need to divide the concentration by 2^2 (since 2 hours is two half-lives) in order to get the concentration at the earlier time.

Mathematically, this can be represented as:

50 ng/dL ÷ 2^2 = 50 ng/dL ÷ 4 = 12.5 ng/dL.

Therefore, the concentration of cocaine in the deceased person's blood two hours earlier would have been approximately 12.5 ng/dL.